# More geometry

• Nov 28th 2005, 05:53 AM
x-disturbed-x
More geometry
A and B are points of intersection of y=x^2-2x and y=2-3x
simultaneous equation solved = x^2 + x -2 = 0

a) Find the coordinates of A and B, given that the x coordinate of B is greater than that of A.

The line L is perpendicular to y=2-3x and passes through B.
b) Find an equation of L

The line M is parallel to y=5-3x and passes through A
c) Find an equation of M

The lime M crosses the y-axis at point P. The line L crosses the x-axis at point Q.
d) Find the distance of PQ, giving your answer in square root form where a and b are integers and b is prime. (Hint - use pythagoras' theorem)
• Dec 1st 2005, 02:48 AM
CaptainBlack
Quote:

Originally Posted by x-disturbed-x
A and B are points of intersection of y=x^2-2x and y=2-3x
simultaneous equation solved = x^2 + x -2 = 0

a) Find the coordinates of A and B, given that the x coordinate of B is greater than that of A.

You have identified correctly that the x co-ordinatea of the
points of intersection of A and B are the solutions of

\$\displaystyle x^2+x-2\ =\ 0\$

so now find those solutions, let's call them \$\displaystyle x_1\$ and \$\displaystyle x_2\$, and we
will assume that \$\displaystyle x_1\ <\ x_2\$, (they are not equal as the LHS
of the equation is not the square of a single linear factor).

Then from what we are told we know that the
x co-ordinate of A is \$\displaystyle x_1\$, and the x co-ordinate of B
is \$\displaystyle x_2\$.

To find the y co-ordinates of A and B respectivly we now substitute
\$\displaystyle x_1\$ and \$\displaystyle x_2\$ into

\$\displaystyle y\ =\ 2-3x\$

RonL
• Dec 1st 2005, 05:30 AM
CaptainBlack
Quote:

Originally Posted by x-disturbed-x
The line L is perpendicular to y=2-3x and passes through B.
b) Find an equation of L

The gradient of any line perpendicular to the line:

\$\displaystyle y\ =\ m.x\ +\ c\$

is \$\displaystyle -1/m\$.

So you have sufficient information to determine the slope
of L. Also you now know the co-ordinates of B (as you
will have found them in part a), you can find in intercept
for L by plugging the co-ordinates of B into into the
general equation of a line with the required slope.

RonL
• Dec 1st 2005, 05:35 AM
CaptainBlack
[QUOTE=x-disturbed-x]The line M is parallel to y=5-3x and passes through A
c) Find an equation of M

This is like part b, except that the slope of M can be
found using the result that the slopes of parallel lines are equal.

Then proceed as in part b.

RonL
• Dec 1st 2005, 08:07 AM
CaptainBlack
Quote:

Originally Posted by x-disturbed-x

The lime M crosses the y-axis at point P. The line L crosses the x-axis at point Q.
d) Find the distance of PQ, giving your answer in square root form where a and b are integers and b is prime. (Hint - use pythagoras' theorem)

You should by now have the equations for M and L, let these be:

\$\displaystyle \mbox{L: }\ \ y\ =\ m_L.x\ +\ c_L\$

\$\displaystyle \mbox{M: }\ y\ =\ m_M.x\ +\ c_M\$

M crosses the y-axis when \$\displaystyle x\ =\ 0\$, which is at

\$\displaystyle y\ =\ c_M\$

so \$\displaystyle P\$ is the point \$\displaystyle (0,c_M)\$.

Similarly L crosses the x-axis when \$\displaystyle y\ =\ 0\$, which is at

\$\displaystyle x\ =\ -c_L/m_M\$

so \$\displaystyle Q\$ is the point \$\displaystyle (-c_L/m_L,0)\$.

So now you should know where \$\displaystyle P\$ and \$\displaystyle Q\$ are
and so you can find the distance between them in the usual manner.

RonL